A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $\frac{1}{3}$ of area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_{1}$ . When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ .The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option, ignoring edge effects.

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$\frac{E_{1}}{E_{2}} = 1$

$\frac{E_{1}}{E_{2}} = \frac{1}{K}$

$\frac{Q_{1}}{Q_{2}} = \frac{3}{K}$

$\frac{C}{C_{1}} = \frac{2 + K}{K}$

answer is A, D.

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Detailed Solution

$C = C_{1} + C_{2}; C_{1} = \frac{k ε_{0} \frac{A}{3}}{d}; C_{2} = \frac{\frac{ε_{0} 2 A}{3}}{d}; \Rightarrow C = \frac{(K + 2) ε_{0} A}{3 d} \Rightarrow \frac{C}{C_{1}} = \frac{K + 2}{K}$

Also, $E_{1} = E_{2} = \frac{V}{d}$ , where $V$ is potential difference between the plates.

$\frac{Q_{1}}{Q_{2}} = \frac{C_{1} V}{C_{2} V} = \frac{C_{1}}{C_{2}} = \frac{K}{2}$

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